Find the value of 'a' so that the modulus of the complex number is 1300 and belong to the 1st quadrant

Question

Given: $z=(24-10i)·(a+40i)$ find $a$ so that $|z|=1300$

I guess I have to $\sqrt{something^2+something^2}=1300$ and then isolate $a$ but I don't know how to put $z$ in $a+bi$ form.

Answer 1

Hint: Use the fact that $|z.w|=|z|.|w|$.

Answer 2

To put $z$ in the form $x+yi$ you can expand the brackets:

$z=24a+400+(960-10a)i$

Then $\sqrt{(24a+400)^2+(960-10a)^2}=1300$.

But it will be quicker to use José Carlos Santos's hint.